4
$\begingroup$

The sequence of Genocchi numbers
${({G_{2n}})_{n \ge 0}}=$ $(0,1,1,3,17,155,2073,...)$

can be defined by the generating function $z\frac{{1 - {e^z}}}{{1 + {e^z}}} = \sum {{{( - 1)}^n}{G_{2n}}\frac{{{z^{2n}}}}{{(2n)!}}} .$

Many different q-analogs of these numbers have been studied. Does anyone know if the following q-analog ${({G_{2n}(q)})_{n \ge 0}}$ is known? It is intimately related with q-Chebyshev polynomials.

Let $(a;q)_n=(1-a)(1-qa) \cdots (1-q^{n-1}a)$, $[n]=1+q+\cdots+q^{n-1}$ and $[n]!=[1][2] \cdots[n].$

The q-analog can defined by the generating function

$\sum\limits_{n \ge 1} {\frac{{{{( - 1)}^{n - 1}}{G_{2n}}(q){{( - q;q)}_{2n - 1}}}}{{[2n]!}}} {z^{2n}} = $

$\sum\limits_{n \ge 1} {\frac{{{{( - q;q)}_{2n - 1}}}}{{[2n]!}}} {z^{2n}} $ divided by

$\sum\limits_{n \ge 0} {\frac{{{{( - q;q)}_{2n}}}}{{[2n + 1]!}}} {z^{2n}}.$

$\endgroup$
3
  • $\begingroup$ Is this the same as the q-analog you get by rewriting $\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k {n \choose 2k}G_{2n-k} = 0$ to a relation involving q-binomial coefficients? $\endgroup$ Jul 14, 2012 at 22:41
  • 1
    $\begingroup$ The corresponding Seidel identity is $$\sum{(-1)^k}q^{\binom{2k}{2}} {n\brack{2k}}{{(-q^{n-2k+1};q)_{2k}}}/ {{(-q^{2n-2k};q)_{2k}}}{G_{2n-2k}}(q) =[n=1].$$ $\endgroup$ Jul 15, 2012 at 6:45
  • 1
    $\begingroup$ In the mean time I have seen that these q-Genocchi numbers are related to the usual $q-$tangent numbers ${T_{2n - 1}}(q)$ by ${(- q;q)_ {2n - 1}} {G_{2n}}(q) = [2n] {T_{2n - 1}}(q).$ $\endgroup$ Jul 20, 2012 at 8:57

1 Answer 1

1
$\begingroup$

Different definitions of the q-Genocchi numbers and polynomials have been studied by many mathematicians for a long time, for instance: T. Kim, L. C. Jang, C. S. Ryoo, Y. Simsek, S. Araci, H. Jolany,...etc. So that, the readers can refer to the link: http://www.hindawi.com/journals/jfsa/2012/214961/

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.